The coordination geometry around and hybridization of N and B atoms in a 1:1 complex of NH₃ and BF₃ ${\mathrm{is}}_{}$:

(1) N: tetrahedral, sp³;  B: tetrahedral, sp³

(2) N: pyramidal, sp³; B: pyramidal, sp³

(3) N: pyramidal, sp³; B: planar, sp²

(4) N: pyramidal, sp³; B: tetrahedral, sp³

Assuming $2s-2p$ mixing is NOT operative, the paramagnetic species among the following is :

(1) $B{e}_2$

(2) $B_2$

(3) $C_2$

(4) $N_2$

When $O_2$ is adsorbed on a metallic surface, electron transfer occurs from the ,metal to $O_2$. The wrong statements(s) regarding this adsorption is (are): 

(1) $O_2$ is physisorbed 

(2) Heat is released

(3) Occupancy of $\pi *2p$ of $O_2$ is increased 

(4) Bond length of $O_2$ is increased

Born Haber cycle is used to determine:

(1) lattice energy 

(2) electron affinity 

(3) ionization energy 

(4) either of them 

The molecule having the smallest bond angle is-

Select the most ionic and most covalent compounds respectively from the following.

$Cr{O}_5,M{n}_2{O}_7,PbO, P_4{O}_{10}, Sn{O}_2$

(A) $Cr{O}_5, M{n}_2{O}_7$                                (B)  $PbO, M{n}_2{O}_7$

(C) $Cr{O}_5, P_4{O}_{10}$                                 (D) $Sn{O}_2, Cr{O}_5$

In which of the following sets the central atom of each member involves $s{p}^3$ hybridisation?

(A) $I{O_4}^{‐}IC{l_4}^{‐}and I{F_4}^{+}$                                      (B) $Xe{O}_3, Xe{O}_4, Xe{F}_4$

(C) $S{O}_3,\hspace{0.17em}S{O_3}^{2‐}, S{O_4}^{2‐}$                                    (D) $PC{l_4}^{+}, B{F_4}^{‐}, Cl{O_4}^{‐}$

In which of the following groups all the members have linear shape?

(1) $N{O}_2, {N_3}^{‐}, IC{l_2}^{+}$                          

(2) ${N_3}^{‐}, {I_3}^{‐}, N{O_2}^{+}$

(3) $Xe{F}_2, C_2{H}_2, S{O}_2$                        

(4) $C{O}_2, BeC{l}_2, SnC{l}_2$

Which reaction involves a change in the HYBRIDIZATION for the under lined geometry?

(A) $\underline{B}{F}_3+ F\to B{F_4}^{-}$                           (B) $\underline{N}{H}_3 + H^{+}\to \underline{N}{H_4}^{+}$

(C) $2\underline{S}{O}^2+O_2\to 2\underline{S}{O}_3$                        (D) $H_2\underline{O} + H^{+}\to H_3{\underline{O}}^{+}$

The ${\mathrm{BF}}_3$ is a planar molecule whereas ${\mathrm{NF}}_3$ is pyramidal because:

1. $\mathrm{B}-\mathrm{F}$ bond is more polar than $\mathrm{N}-\mathrm{F}$ bond.

2. Boron atom is bigger than nitrogen atom.

3. Nitrogen is more electronegative than boron.

4. ${\mathrm{BF}}_3$ has no lone pair but ${\mathrm{NF}}_3$ has a lone pair of electrons.